# Statics-Two wires holding traffic light

I dont know what i did wrong

Find the tension in the two wires supporting the traffic light
Mass of light= 33 kg
θ of FT1= 53°
θ of FT2=37°

Ʃfx=0
Ft1cos53-ft2cos37=0
ft1cos53=ft2cos37

Ʃfy=mg
ft1sin53-ft2sin37=mg
ft1sin53=mg+ft2sin37
ft1=(ft2sin37+mg)/sin53
ft1=1.33ft2

1.33ft2=(ft2sin37+mg)/sin53
1.33ft2=.72ft2+mg
.61ft2=323.73 N
Ft2=530.7n

ft1cos53=ft2cos37
ft1cos53=423.84n
ft1=704.26n

## The Attempt at a Solution

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PhanthomJay
Homework Helper
Gold Member
Tension forces always pull away from the objects on which they act. In the horizontal direction, your equation is correct. Not so in the vertical direction.

This is what I got:
TLsin(53o) + TRsin(37o) - 294 N= 0 TRcos(37o) - TLcos(53o) = 0
TR = TLcos(53o)/cos(37o)

TLsin(53o) + TRsin(37o) - 294N= 0
TLsin(53o) + {TLcos(53o)/cos(37o)}sin(37o) - 294 = 0
TLsin(53o) + TLcos(53o)tan(37o) = 294N
TL(sin(53o) + cos(53o)tan(37o)) = 294N
TL = (294 N)/(sin(53o) + cos(53o)tan(37o)) = 234.79884N = 230N

TR = TLcos(53o)/cos(37o)
TR = (234.79884 N)cos(53o)/cos(37o) = 176.9336 N = 180N

Left wire has 230N of tension
Right wire has 180N of tension

I don't know where you got the 294N from? 33kg * 9.8 = 323N. g is 9.8, not 8.9!

You can check your answer easily also. 230sin(53) + 180sin(37) = 292N, which is close enough (since you rounded your answers) to 294.

Well dangit... I guess I got that one wrong then!

PhantomJay is correct

Hope this helps.